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Counting the number of characters

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# 1  
Old 02-20-2012
Counting the number of characters
Hi all,

Can someone help me in getting the following o/p

I/p:
1157,306413,R,2009,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 ,0,0,0,0,0,0,0,0,0,0,0,,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,,0,0,0,0,0,0,0,0,0,5,405,0,0,102,102,102,102,102,102,102,103,102,102,102 ,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102 ,102,102,102,,102,102,102,102,102,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

O/p:
1157,306413,R,2009,113,

The First four fields remains the same , after that come the number of Zero or Null characters before the first non-zero or non-null charqacter , in the example given above it is "5" and the count is 113.

Would be really helpful if some one can help me by giving a statement (preferably awk )for this.

Many Thanks in advance
Sri


# 2  
Old 02-20-2012
Try:
Code:
awk -F, '{for(i=5;!$i>0 && i<=NF;i++);NF=5;$5=i-5}1' OFS=, infile

# 3  
Old 02-20-2012
Thank you Scrutinizer,

Can you please explain me the logic , because this is just one step of my requirement i need to do some more steps in this , I would need the o/p of that 5th Coloumn , My final o/p should look something like this

I/p:
1157,306413,R,2009,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 ,0,0,0,0,0,0,0,0,0,0,0,,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,,0,0,0,0,0,0,0,0,0,5,405,0,0,102,102,102,102,102,102,102,103,102,102,102 ,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102 ,102,102,102,,102,102,102,102,102,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,

O/p:
1157,306413,R,2009,113,5,405,0,0,102,102,102,102,102,102,102,103,102,102,102,102,102,102,102,102,102 ,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,102,,102,102,10 2,102,102,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0 ,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0, 0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,0,,0,0,0,0,0,0,0,0,0

I need to remove the zeros or null before the non zero character and append it at the end so that my coloumn count remains the same

Many thanks in advance

# 4  
Old 02-20-2012
You means like this?
Code:
awk -F, '{s=x;for(i=5;!($i>0) && i<=NF;i++)s=s OFS $i;sub(s,x);$0=$0s}1' OFS=, infile

-F,Use comma as input field separator
s=xmake s equal to "" (the same value as x, which is initialized)
for(i=5;!($i>0) && i<=NF;i++)s=s OFS $iadd each field together with the output field separator (OFS) to the string s that does not contain a positive value (and stop if the last field is reached)
sub(s,x)replace all those null fields contained in variable s with an empty in the record ($0)
$0=$0spaste them at the end of the record ($0)
1print the record
OFS=,Use comma as output field separator
# 5  
Old 02-20-2012
Sorry,

That is not giving the count of zeros and nulls that was coming before the non null value,

Can you please give me a statement to delete these fields , probably we can append it as the last step of the script
# 6  
Old 02-20-2012
Just combine the two
Code:
awk -F, '{s=x;for(i=5;!($i>0) && i<=NF;i++)s=s OFS $i;sub(s,OFS i-5);$0=$0s}1' OFS=, infile

What had you tried yourself?
Last edited by Scrutinizer; 02-20-2012 at 09:30 AM..
# 7  
Old 02-20-2012
It is not working ,

I was trying to get the o/p of the first command you have given save it as a file , Then remove the coloumns between the 6th coloumn and the coloumn we get from the 5th coloumn , then append this to the last of the file.
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